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0.45 g an acid (mol wt.=90) required 20 ml of 0.5 N KOH for complete neutralization. Basicity of acid is

Answer»

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Solution :Normality`=N=(W_(B)xx1000)/(EQ." "wtxxV)`
`therefore"Eq. WT"=(0.45xx1000)/(0.5xx20)=45`
`therefore"BASICITY"=("Molec. Wt")/("Eq. wt")=(90)/(45)=2`


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